MT100.22 ASSIGNMENT EIGHT—SOLUTIONS
3.5.4 Calculate the second and third derivatives of y = 4t 3 − 9t2 + 7.
Solution. First compute the first derivative:
y 0 = 12t2 − 18t
Then the second derivative is just the derivative of the first derivative:
y 00 = 24t − 18
And the third derivative is the derivative of the second derivative:
y 000 = 24
3.5.28 Compute f 000 (0) where f (x) = 4ex − x3 .
Solution. Note that
f 0 (x) = 4ex − 3x2
f 00 (x) = 4ex − 6x
and so
f 000 (x) = 4ex − 6
f 000 (0) = 4e0 − 6 = 4 − 6 = −2
3.6.2 Find an equation of the tangent line to the graph of f (x) = cos(x) at x = π3 .
Solution. To find an equation of a tangent line, we must find
the slope and a point. The
slope of the tangent line to the graph of f at x = π3 is f 0 π3 . Since f 0 (x) = − sin(x), the
slope of the tangent line is
π √3
0 π
= − sin
=
f
.
3
3
2
The point on the line is π3 , f π3 . In this case, f π3 = cos π3 = 12 . An equation of
the tangent line, then, is
√ 1
3
π
y− =
x−
2
2
3
x
3.6.18 Find the derivative of f (x) = tan(x) .
Solution. Here, we require the quotient rule:
f 0 (x) =
tan(x)(1) − x sec2 (x)
.
tan2 (x)
With a little bit of (optional) manipulation, we see that
f 0 (x) =
sin(x) cos(x) − x
sin2 (x)
= cot(x) − x csc2 (x)
3.7.10 Find the derivative of y = sin5 (x).
Solution. Observe that y = sin5 (x) = (sin(x))5 . Thus, by the chain rule,
y 0 = 5 (sin(x))4 (cos(x)) = 5 sin4 (x) cos(x).
1
3.7.42 Find the derivative of f (x) = sin(cos(sin(x))).
Solution. We must repeatedly apply the chain rule:
d
df
= cos(cos(sin(x)))
(cos(sin(x)))
dx
dx
d
= cos(cos(sin(x))) (− sin(sin(x)))
(sin(x))
dx
= − cos(cos(sin(x))) sin(sin(x)) cos(x).
3.7.68 Find the derivative of f (x) = e e .
Solution. Again, we must repeatedly apply the chain rule:
df
x d
= ee
(ex )
dx
dx
x
= e e ex
x
= ex+e
x
3.7.92 The price (in dollars) of a computer component is P = 2C − 18C −1 , where C is the
manufacturer’s cost to produce it. Assume that cost at time t (in years) is C = 9 + 3t −1
and determine the rate of change of price with respect to time at t = 3.
Solution. We are asked to compute dP
dt . Noting that we have P in terms of C and C in
terms of t, the chain rule implies that
dP
dP dC
=
dt
dC dt
Now,
dC
dP
= 2 + 18C −2 and
= −3t−2
dC
dt
and so
dP
= 2 + 18C −2 −3t−2 .
dt
To finish the job, simply find C when t = 3 and substitute. That is, C(3) = 9+3(3) −1 = 10,
so
2.18
dP
= 2 + 18(10)−2 −3(3)−2 = −
= −0.726
dt
3
2